The lifespans of porcupines in a particular zoo are normally distributed. The average porcupine lives $16.8$ years; the standard deviation is $1.9$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a porcupine living less than $13$ years.
Solution: $16.8$ $14.9$ $18.7$ $13$ $20.6$ $11.1$ $22.5$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $16.8$ years. We know the standard deviation is $1.9$ years, so one standard deviation below the mean is $14.9$ years and one standard deviation above the mean is $18.7$ years. Two standard deviations below the mean is $13$ years and two standard deviations above the mean is $20.6$ years. Three standard deviations below the mean is $11.1$ years and three standard deviations above the mean is $22.5$ years. We are interested in the probability of a porcupine living less than $13$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the porcupines will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the porcupines will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $13$ years and the other half $({2.5\%})$ will live longer than $20.6$ years. The probability of a particular porcupine living less than $13$ years is ${2.5\%}$.